Areas Related to Circles

1. What this chapter is really about

In earlier classes you learnt the two master formulae for a circle of radius $r$:

This chapter does not add a single brand-new "big" idea. Instead it asks: what if we only want a slice of the circle, not the whole thing? A clock's minute hand sweeps part of a circle in $5$ minutes; a windscreen wiper cleans part of a circle; a tethered horse grazes part of a circle. To handle all of these we only need two new sub-shapes — the sector and the segment — and the single trick of taking a fraction of the whole circle.

Throughout, $\pi$ (pi) is the constant $\approx 3.14159\dots$ Boards usually tell you to use $\pi=\dfrac{22}{7}$ (handy when the radius is a multiple of $7$) or $\pi=3.14$. Always use the value the question specifies.

2. Sector and segment — the two key shapes

Sector: the part of the circular region enclosed by two radii and the arc between them — like a slice of pizza. In the figure the shaded slice $OAPB$ is a sector, and the angle $\angle AOB$ at the centre is called the angle of the sector.

Segment: the part of the circular region enclosed between a chord and the arc standing on it — like the bit you slice off the top of an orange. The chord $AB$ cuts the circle into two segments.

• The smaller piece is the minor sector/segment; the larger piece is the major sector/segment.
• The angle of the major sector $=360^{\circ}-\angle AOB$ (the rest of the full turn).
• Convention: if a question just says "sector" or "segment", it means the minor one, unless told otherwise.

The whole circle itself can be thought of as one giant sector whose angle is the full turn $360^{\circ}$. That observation is the key to deriving every formula below.

3. Area of a sector — the $\dfrac{\theta}{360}$ idea

Let the sector $OAPB$ have radius $r$ and central angle $\theta$ (in degrees). We use the Unitary Method, comparing the slice to the full circle:

• An angle of $360^{\circ}$ at the centre gives the full area $\pi r^{2}$.
• So an angle of $1^{\circ}$ gives $\dfrac{\pi r^{2}}{360}$.
• Therefore an angle of $\theta$ gives $\dfrac{\pi r^{2}}{360}\times\theta$.

In words: a sector is simply the fraction $\dfrac{\theta}{360}$ of the whole circle's area. If $\theta=90^{\circ}$ you get a quarter (quadrant); if $\theta=180^{\circ}$ you get a semicircle.

4. Length of an arc

The same fraction idea gives the length of the curved boundary (the arc $APB$) of the sector. The whole circumference is $2\pi r$, so:

Notice the neat link: a sector of angle $\theta$ has area $\dfrac{\theta}{360}\pi r^{2}$ and arc $\dfrac{\theta}{360}\,2\pi r$. There is also a tidy relation between them:

The perimeter of a sector is the arc plus the two straight radii: $\text{arc}+2r=\dfrac{\theta}{360}\,2\pi r+2r$.

5. Area of a segment

A segment is a sector with its triangular part removed. Look at the minor segment $APB$ cut off by chord $AB$. Join the centre $O$ to $A$ and $B$ to form triangle $OAB$. Then:

To get the area of $\triangle OAB$, drop a perpendicular $OM$ from the centre to the chord; it bisects both the chord and the angle. Then with $\angle AOM=\dfrac{\theta}{2}$:

(The shortcut $\tfrac12 r^{2}\sin\theta$ is handy, but boards expect the full $\tfrac12\times\text{base}\times\text{height}$ working too.)

6. Major sector and major segment

The "leftover" big piece is found by subtracting the minor piece from the whole circle:

Equivalently the major sector $=\dfrac{360-\theta}{360}\times\pi r^{2}$, using its own angle $360-\theta$.

7. NCERT Example 1 — sector and major sector

8. NCERT Example 2 — area of a segment

9. NCERT Exercise 11.1 — fully solved (Q1–Q7)

Q1. Sector, $r=6$ cm, $\theta=60^{\circ}$. Area $=\dfrac{60}{360}\times\dfrac{22}{7}\times6\times6=\dfrac16\times\dfrac{22}{7}\times36=\dfrac{132}{7}\approx 18.86\ \text{cm}^{2}.$

Q2. Quadrant ($\theta=90^{\circ}$), circumference $=22$ cm. First $2\pi r=22\Rightarrow r=\dfrac{22}{2}\times\dfrac{7}{22}=\dfrac72$ cm. Quadrant area $=\dfrac{90}{360}\times\dfrac{22}{7}\times\left(\dfrac72\right)^{2}=\dfrac14\times\dfrac{22}{7}\times\dfrac{49}{4}=\dfrac{77}{8}\approx 9.625\ \text{cm}^{2}.$

Q3. Minute hand $r=14$ cm. In $5$ minutes it turns $\dfrac{5}{60}\times360^{\circ}=30^{\circ}$. Area swept $=\dfrac{30}{360}\times\dfrac{22}{7}\times14\times14=\dfrac{1}{12}\times\dfrac{22}{7}\times196=\dfrac{154}{3}\approx 51.33\ \text{cm}^{2}.$

Q4. $r=10$ cm, right angle $\theta=90^{\circ}$, $\pi=3.14$.

• Sector area $=\dfrac{90}{360}\times3.14\times100=78.5\ \text{cm}^{2}.$ Triangle $OAB$ is right-angled at $O$ with legs $=r=10$: area $=\dfrac12\times10\times10=50\ \text{cm}^{2}.$
• (i) Minor segment $=78.5-50=28.5\ \text{cm}^{2}.$
• (ii) Major sector $=\pi r^{2}-\text{minor sector}=3.14\times100-78.5=314-78.5=235.5\ \text{cm}^{2}.$

Q5. $r=21$ cm, $\theta=60^{\circ}$, $\pi=\dfrac{22}{7}$.

• (i) Arc $=\dfrac{60}{360}\times2\times\dfrac{22}{7}\times21=\dfrac16\times132=22\ \text{cm}.$
• (ii) Sector $=\dfrac{60}{360}\times\dfrac{22}{7}\times441=\dfrac16\times1386=231\ \text{cm}^{2}.$
• (iii) Segment $=$ sector $-\triangle OAB$. Here $\triangle OAB$ has $OA=OB=21$ and included angle $60^{\circ}$, so it is equilateral: area $=\dfrac{\sqrt3}{4}\times21^{2}=\dfrac{441\sqrt3}{4}\approx 190.95\ \text{cm}^{2}.$ Segment $=231-\dfrac{441\sqrt3}{4}\approx 40.05\ \text{cm}^{2}.$

Q6. $r=15$ cm, $\theta=60^{\circ}$, $\pi=3.14$, $\sqrt3=1.73$. Sector $=\dfrac{60}{360}\times3.14\times225=\dfrac16\times706.5=117.75\ \text{cm}^{2}.$ Triangle equilateral: $\dfrac{\sqrt3}{4}\times225=\dfrac{1.73}{4}\times225\approx 97.31\ \text{cm}^{2}.$

• Minor segment $=117.75-97.31=20.44\ \text{cm}^{2}.$
• Major segment $=\pi r^{2}-\text{minor segment}=3.14\times225-20.44=706.5-20.44=686.06\ \text{cm}^{2}.$

Q7. $r=12$ cm, $\theta=120^{\circ}$, $\pi=3.14$, $\sqrt3=1.73$. Sector $=\dfrac{120}{360}\times3.14\times144=\dfrac13\times452.16=150.72\ \text{cm}^{2}.$ Triangle $OAB$: area $=\dfrac12 r^{2}\sin120^{\circ}=\dfrac12\times144\times\dfrac{\sqrt3}{2}=36\sqrt3=36\times1.73=62.28\ \text{cm}^{2}.$ Segment $=150.72-62.28=88.44\ \text{cm}^{2}.$

10. NCERT Exercise 11.1 — fully solved (Q8–Q14)

Q8. Horse tied at a corner of a $15$ m square with a rope, $\pi=3.14$.

• (i) At a corner the rope sweeps a quarter circle ($90^{\circ}$). With rope $5$ m: grazing area $=\dfrac{90}{360}\times3.14\times5^{2}=\dfrac14\times78.5=19.625\ \text{m}^{2}.$
• (ii) With rope $10$ m: area $=\dfrac14\times3.14\times100=78.5\ \text{m}^{2}.$ Increase $=78.5-19.625=58.875\ \text{m}^{2}.$ (The $10$ m rope still fits within the $15$ m side, so it stays a quarter circle.)

Q9. Brooch: circle of diameter $35$ mm, so $r=\dfrac{35}{2}$ mm; $5$ diameters divide it into $10$ equal sectors.

• (i) Wire $=$ circumference $+5$ diameters $=2\pi r+5d=\dfrac{22}{7}\times35+5\times35=110+175=285\ \text{mm}.$
• (ii) Each sector angle $=\dfrac{360}{10}=36^{\circ}$. Area $=\dfrac{36}{360}\times\dfrac{22}{7}\times\left(\dfrac{35}{2}\right)^{2}=\dfrac{1}{10}\times\dfrac{22}{7}\times\dfrac{1225}{4}=\dfrac{385}{4}=96.25\ \text{mm}^{2}.$

Q10. Umbrella: $8$ equally spaced ribs, $r=45$ cm. Area between two consecutive ribs $=$ one of $8$ equal sectors $=\dfrac{360/8}{360}\times\pi r^{2}=\dfrac18\times\dfrac{22}{7}\times2025=\dfrac{22275}{28}\approx 795.54\ \text{cm}^{2}.$

Q11. Wiper blade $r=25$ cm, $\theta=115^{\circ}$, two non-overlapping wipers. One sweep $=\dfrac{115}{360}\times\dfrac{22}{7}\times625$. Total (two blades) $=2\times\dfrac{115}{360}\times\dfrac{22}{7}\times625=\dfrac{158125}{126}\approx 1254.96\ \text{cm}^{2}.$

Q12. Lighthouse: $\theta=80^{\circ}$, $r=16.5$ km, $\pi=3.14$. Warned area $=\dfrac{80}{360}\times3.14\times16.5^{2}=\dfrac{2}{9}\times3.14\times272.25\approx 189.97\ \text{km}^{2}.$

Q13. Round table cover, $6$ equal designs (segments), $r=28$ cm, $\sqrt3=1.7$. Each design subtends $\dfrac{360}{6}=60^{\circ}$. Area of one segment $=$ sector $-$ equilateral triangle $=\dfrac{60}{360}\times\dfrac{22}{7}\times784-\dfrac{\sqrt3}{4}\times784=410.67-333.2=77.47\ \text{cm}^{2}.$ Six designs $=6\times77.47=464.8\ \text{cm}^{2}.$ Cost $=464.8\times0.35\approx \text{₹}162.68.$

Q14. Area of a sector of angle $p$ degrees, radius $R$. The standard formula is $\dfrac{p}{360}\times\pi R^{2}$, which equals $\dfrac{p}{720}\times2\pi R^{2}$. So the correct option is (D) $\dfrac{p}{720}\times2\pi R^{2}.$

11. Common mistakes to avoid

• Confusing arc length ($\dfrac{\theta}{360}\times2\pi r$) with sector area ($\dfrac{\theta}{360}\times\pi r^{2}$) — one uses $2\pi r$, the other $\pi r^{2}$.
• Forgetting to subtract the triangle when asked for a segment — segment $\neq$ sector.
• Using the wrong value of $\pi$ — read whether the question says $\dfrac{22}{7}$ or $3.14$.
• Mixing units: convert minutes to a turn angle ($1$ min $=6^{\circ}$, since $\dfrac{360}{60}=6$) before using the sector formula.
• For the major piece, either subtract the minor from $\pi r^{2}$ or use angle $360-\theta$ — don't do both.
• Taking the triangle area as $\tfrac12 r^{2}$ when the angle is not $90^{\circ}$ — only a right-angle sector gives a right triangle with legs $r,r$.

12. Quick revision checklist

• Circle: circumference $2\pi r$, area $\pi r^{2}$.
• Sector area $=\dfrac{\theta}{360}\pi r^{2}$; arc length $=\dfrac{\theta}{360}\,2\pi r$.
• Segment area $=$ sector area $-$ area of $\triangle OAB$.
• $\triangle OAB=\dfrac12 r^{2}\sin\theta$ (or $\dfrac12\times$ base $\times$ height with $OM=r\cos\tfrac\theta2$).
• Major piece $=\pi r^{2}-$ minor piece.
• Quadrant $=\tfrac14$ circle ($\theta=90^{\circ}$); semicircle $=\tfrac12$ circle ($\theta=180^{\circ}$).
• Clock: minute hand turns $6^{\circ}$ per minute, hour hand $0.5^{\circ}$ per minute.